Answer
$$-\frac{1}{4}(\ln x)^{-4}+c.$$
Work Step by Step
Let $u=\ln x$, then $du=\frac{dx}{x}$. Now, we have
$$\int \frac{dx}{x(\ln x)^5}=\int \frac{du}{u^5}=-\frac{1}{4}u^{-4}+c\\
=-\frac{1}{4}(\ln x)^{-4}+c.$$
Calculus (3rd Edition)
by Rogawski, Jon; Adams, Colin
Published by
W. H. Freeman
ISBN 10:
1464125260
ISBN 13:
978-1-46412-526-3
Chapter 7 - Exponential Functions - 7.8 Inverse Trigonometric Functions - Exercises - Page 375: 98
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